MIE 2015/Day 2/Problem 2

Problem 2

Let the functions $f_n$ , for $n\in\{0,1,2,3,...\}$ , such that $f_0(x)=\frac{1}{1-x}$ and $f_n(x)=f_0(f_{n-1}(x))$ , for every $n\geq1$ .

Compute $f_{2016}(2016)$ .

First, let see the case $n=1$

$f_1(x)=f_0(f_{1-1}(x))$

$f_1(x)=\frac{1}{1-\frac{1}{1-x}}$

$f_1(x)=\frac{x-1}{x}$

Now, when $n=2$

$f_2(x)=f_0(f_1(x))$

$f_2(x)=\frac{1}{1-\frac{x-1}{x}}$

$f_2(x)=x$

Now, when $n=3$

$f_3(x)=\frac{1}{1-x}$

At this point it's easy to see the pattern. So, we just find the remainder of 2016 by 3.

$f_{2016}(x)=f_0(x)$

$f_{2016}(2016)=\frac{1}{1-2016}$

$f_{2016}(2016)=-\frac{1}{2015}$